What is the extraneous solution to these equations? $\dfrac{x^2 + 42}{x - 10} = \dfrac{x + 132}{x - 10}$
Answer: Multiply both sides by $x - 10$ $ \dfrac{x^2 + 42}{x - 10} (x - 10) = \dfrac{x + 132}{x - 10} (x - 10)$ $ x^2 + 42 = x + 132$ Subtract $x + 132$ from both sides: $ x^2 + 42 - (x + 132) = x + 132 - (x + 132)$ $ x^2 + 42 - x - 132 = 0$ $ x^2 - 90 - x = 0$ Factor the expression: $ (x + 9)(x - 10) = 0$ Therefore $x = -9$ or $x = 10$ At $x = 10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 10$, it is an extraneous solution.